The Fourier Transform
As we have seen, any (sufficiently smooth) function f (t) that is periodic can be built out of sin’s and
cos’s. We have also seen that complex exponentials may be used in place of sin’s and cos’s. We shall now
use complex exponentials because they lead to less writing and simpler computations, but yet can easily be
converted into sin’s and cos’s. If f (t) has period 2ℓ, its (complex) Fourier series expansion is
Z ℓ
∞
X
π
ik πℓ t
1
ck e
f (t) =
with
ck = 2ℓ
(1)
f (t)e−ik ℓ t dt
−ℓ
k=−∞
ik πℓ t
π
Not surprisingly, each term ck e
in this expansion also has period 2ℓ, because ck eik ℓ (t+2ℓ) =
π
π
ck eik ℓ t ei2kπ = ck eik ℓ t . We now develop an expansion for non-periodic functions, by allowing complex
exponentials (or equivalently sin’s and cos’s) of all possible periods, not just 2ℓ, for some fixed ℓ. So, from
now on, do not assume that f (t) is periodic.
For simplicity we’ll only develop the expansions for functions that are zero for all sufficiently large |t|.
With a little more work, one can show that our conclusions apply to a much broader class of functions. Let
L > 0 be sufficiently large that f (t) = 0 for all |t| ≥ L. We can get a Fourier series expansion for the part
of f (t) with −L < t < L by using the periodic extension trick. Define FL (t) to be the unique function
determined by the requirements that
i) FL (t) = f (t) for − L < t ≤ L
ii) FL (t) is periodic of period 2L
Then, for −L < t < L,
f (t) = FL (t) =
∞
X
π
ck (L)eik L t
where
ck (L) =
1
2L
k=−∞
Z
L
π
f (t)e−ik L t dt
(2)
−L
If we can somehow take the limit L → ∞, we will get a representation of f that is is valid for all t’s , not
just those in some finite interval −L < t < L. This is exactly what we shall do, by the simple expedient
of interpreting the sum in (2) as a Riemann sum approximation to a certain integral. For each integer k,
π
π
define the k th frequency to be ωk = k L
and denote by ∆ω = L
the spacing, ωk+1 − ωk , between any two
R∞
successive frequencies. Also define fˆ(ω) = −∞ f (t)e−iωt dt. Since f (t) = 0 for all |t| ≥ L
Z L
Z ∞
Z ∞
π
π
1
1
1
1 ˆ
1 ˆ
ck (L) = 2L
f (ωk ) = 2π
f (ωk ) ∆ω
f (t)e−ik L t dt = 2L
f (t)e−i(k L )t dt = 2L
f (t)e−iωk t dt = 2L
−L
−∞
−∞
In this notation,
f (t) = FL (t) =
1
2π
∞
X
fˆ(ωk )eiωk t ∆ω
(3)
k=−∞
for any −L < t < L. As we let L → ∞, the restriction −L < t < L disappears, and the right hand
R∞
1
iωt
ˆ
side converges exactly to the integral 2π
dω. To see this, cut the domain of integration into
−∞ f (ω)e
y
y=
0 ω1 ω2 ω3
c Joel Feldman.
2007. All rights reserved.
∆ω
March 1, 2007
1 ˆ
iωt
2π f (ω)e
ω
The Fourier Transform
1
small slices of width ∆ω and approximate, as in the above figure, the area under the part of the graph of
1 ˆ
1 ˆ
iωt
with ω between ωk and ωk + ∆ω by the area of a rectangle of base ∆ω and height 2π
f (ωk )eiωk t .
2π f (ω)e
R∞
P∞
1
1
iωt
iωk t
ˆ
ˆ
This approximates the integral 2π −∞ f (ω)e
dω by the sum 2π k=−∞ f (ωk )e
∆ω. As L → ∞ the
approximation gets better and better so that the sum approaches the integral. So taking the limit of (3) as
L → ∞ gives
Z ∞
Z ∞
iωt
1
ˆ
ˆ
f (t) = 2π
f (ω)e dω
where
f (ω) =
f (t)e−iωt dt
(4)
−∞
−∞
The function fˆ is called the Fourier transform of f . It is to be thought of as the frequency profile of the
signal f (t).
Example 1 Suppose that a signal gets turned on at t = 0 and then decays exponentially, so that
f (t) =
e−at
0
if t ≥ 0
if t < 0
for some a > 0. The Fourier transform of this signal is
fˆ(ω) =
Z
∞
f (t)e
−iωt
dt =
−∞
Z
∞
e
−at −iωt
e
dt =
0
Z
∞
e
−t(a+iω)
0
∞
1
e−t(a+iω) =
dt =
−(a + iω) 0
a + iω
√
Since a + iω has modulus a2 + ω 2 and phase tan−1 ωa , we have that fˆ(ω) = A(ω)eiφ(ω) with A(ω) =
1
√ 1
and φ(ω) = − tan−1 ωa . The amplitude A(ω) and phase φ(ω) are plotted below.
|a+iω| =
a2 +ω 2
A(ω)
a
a
ω
a
ω
ϕ(ω)
a
Example 2 Suppose that a signal consists of a single rectangular pulse of width 1 and height 1. Let’s
say that it gets turned on at t = − 21 and turned off at t = 12 . The standard name for this “normalized”
rectangular pulse is
1
1
1
rect(t) = 1 if − 2 < t < 2
0 otherwise
t
1
− 21
2
It is also called a normalized boxcar function. The Fourier transform of this signal is
d
=
rect(ω)
c Joel Feldman.
Z
∞
rect(t) e
−iωt
−∞
2007. All rights reserved.
dt =
Z
1/2
e
−1/2
−iωt
1/2
eiω/2 − e−iω/2
e−iωt =
=
dt =
−iω −1/2
iω
March 1, 2007
2
ω
The Fourier Transform
sin ω2
2
d
when ω 6= 0. When ω = 0, rect(0)
=
R 1/2
dt = 1. By l’Hôpital’s rule
−1/2
d
lim rect(ω)
= lim 2
ω→0
ω→0
1
cos ω2
sin ω2
d
= lim 2 2
= 1 = rect(0)
ω→0
ω
1
d
so rect(ω)
is continuous at ω = 0. There is a standard function called “sinc” that is defined(1) by sinc ω =
sin ω
ω
d
d
ω . In this notation rect(ω) = sinc 2 . Here is a graph of rect(ω).
1
d
rect(ω)
−2π
ω
2π
Properties of the Fourier Transform
Linearity
If α and β are any constants and we build a new function h(t) = αf (t) + βg(t) as a linear combination
of two old functions f (t) and g(t), then the Fourier transform of h is
ĥ(ω) =
Z
∞
h(t)e
−iωt
dt =
−∞
Z
∞
−∞
= αfˆ(ω) + βĝ(ω)
αf (t) + βg(t) e−iωt dt = α
Z
∞
f (t)e
−iωt
−∞
dt + β
Z
∞
g(t)e−iωt dt
−∞
(L)
Time Shifting
Suppose that we build a new function h(t) = f (t − t0 ) by time shifting a function f (t) by t0 . The easy
way to check the direction of the shift is to note that if the original signal f (t) has a jump when its argument
t = a, then the new signal h(t) = f (t − t0 ) has a jump when t − t0 = a, which is at t = a + t0 , t0 units to
the right of the original jump.
t0
f (t)
h(t) = f (t − t0 )
a
t
a + t0
The Fourier transform of h is
ĥ(ω) =
Z
=e
∞
−iωt
h(t)e
−∞
Z ∞
−iωt0
dt =
Z
∞
f (t − t0 )e
−iωt
′
fˆ(ω)
dt =
−∞
′
f (t )e
−iωt′
dt = e
−iωt0
Z
∞
′
f (t′ )e−iω(t +t0 ) dt′ where t = t′ + t0 , dt = dt′
−∞
(T)
−∞
(1)
There are actually two different commonly used definitions. The first, which we shall use, is sinc ω =
πω
sinc ω = sinπω
. It is sometimes called the normalized sinc function.
c Joel Feldman.
2007. All rights reserved.
March 1, 2007
sin ω
.
ω
The Fourier Transform
The second is
3
Scaling
If we build a new function h(t) = f
of h is
ĥ(ω) =
Z
∞
h(t)e
−iωt
dt =
−∞
= αfˆ(αω)
Z
∞
f
−∞
t
α
t
α
by scaling time by a factor of α > 0, then the Fourier transform
e
−iωt
dt = α
Z
∞
′
f (t′ )e−iωαt dt′ where t = αt′ , dt = αdt′
−∞
(S)
Example 3 Now consider a signal that consists of a single rectangular pulse of height H, width W and
centre C.
y
H
y = rHWC (t)
C−
W
2
C+
t
W
2
(with rect the normalized rectangular pulse of Example 2) has height
The function rHWC (t) = H rect t−C
W
1
H and jumps when t−C
=
±
,
i.e.
t
=
C ± 12 W and so is the specified signal. By combining properties (L),
W
2
(T) and (S), we can determine the Fourier transform of rHWC (t) = H rect t−C
for any H, C and W . We
W
build it up in three steps.
◦ First we consider the special case in which H = 1 and C = 0. Then we have R1 (t) = rect Wt . So,
according to the scaling property (S) with
f (t) = rect(t),
h(t) = R1 (t) = rect
t
W
=f
t
W
=f
t
α
with α = W
we have
d
R̂1 (ω) = ĥ(ω) = αfˆ(αω) = W rect(W
ω) = W W2ω sin W2ω =
◦ Next we allow a nonzero C too and consider R2 (t) = rect
t−C
W
2
ω
sin W2ω
= R1 (t − C). By (T), with
f (t) = R1 (t), h(t) = R2 (t) = R1 (t − C) = f (t − t0 ), with t0 = C
the Fourier transform is
d Wω
R̂2 (ω) = ĥ(ω) = e−iωt0 fˆ(ω) = e−iωC R̂1 (ω) = W e−iωC rect
◦ Finally, by (L), with α = H and β = 0, the Fourier transform of rHWC = H rect
t−C
W
= HR2 (t) is
Wω
d
r̂HWC (ω) = H R̂2 (ω) = HW e−iωC rect(W
ω) = HW e−iωC W2ω sin W2ω = e−iωC 2H
ω sin 2
Example 4
Now suppose that we have a signal that consists of a number of rectangular pulses of various heights
and widths. Here is an example
c Joel Feldman.
2007. All rights reserved.
March 1, 2007
The Fourier Transform
4
y
2
y = s(t)
1
t
−2
−1
0
1
2
3
4
We can write this signal as a sum of rectangular pulses. As we saw in the last example, rHWC (t) = H rect
is a single rectangular signal with height H, centre C and width W . So
s(t) = s1 (t) + s2 (t) + s3 (t) where
sn (t) = rHWC (t) with
(
t−C
W
H = 2, W = 1, C = −1.5 for n = 1
H = 1, W = 2, C = 1
for n = 2
H = 0.5, W = 2, C = 3
for n = 3
Wω
So, using (L) and r̂HWC (ω) = e−iωC 2H
ω sin 2 , which we derived in Example 3,
ŝ(ω) = ŝ1 (ω) + ŝ2 (ω) + ŝ3 (ω) =
4 i 32 ω
ωe
sin ω2 + ω2 e−iω sin ω + ω1 e−i3ω sin ω
Differentiation
If we build a new function h(t) = f ′ (t) by differentiating an old function f (t), then the Fourier transform
of h is
Z ∞
Z ∞
h(t)e−iωt dt =
ĥ(ω) =
−∞
f ′ (t)e−iωt dt
−∞
Now integrate by parts with u = e−iωt and dv = f ′ (t) dt so that du = −iωe−iωt dt and v = f (t). Assuming
that f (±∞) = 0, this gives
ĥ(ω) =
Z
∞
−∞
Z
∞
u dv = uv −
−∞
∞
−∞
v du = −
Z
∞
f (t) (−iω)e−iωt dt = iω fˆ(ω)
(D)
−∞
Example 5 The differentiation property is going to be useful when we use the Fourier transform to solve
differential equations. As an example, let’s take another look at the RLC circuit
R
L
+
x(t)
−
+
C
y(t)
−
thinking of the voltage x(t) as an input signal and the voltage y(t) as an output signal. If we feed any input
signal x(t) into an RLC circuit, we get an output y(t) which obeys the differential equation
LCy ′′ (t) + RCy ′ (t) + y(t) = x(t)
c Joel Feldman.
2007. All rights reserved.
March 1, 2007
(5)
The Fourier Transform
5
You should be able to quickly derive this equation, which is also (16) in the notes “Fourier Series”, on your
own. Take the Fourier transform of this whole equation and use that
◦ the Fourier transform of y ′ (t) is iω ŷ(ω) and
◦ the Fourier transform of y ′′ (t) is iω times the Fourier transform of y ′ (t) and so is −ω 2 ŷ(ω)
So the Fourier transform of (5) is
−LCω 2 ŷ(ω) + iRCω ŷ(ω) + ŷ(ω) = x̂(ω)
This is trivially solved for
ŷ(ω) =
x̂(ω)
−LCω 2 +iRCω+1
(6)
which exhibits classic resonant behaviour. The circuit acts independently on each frequency ω component
of the signal. The amplitude |ŷ(ω)| of the frequency ω part of the output signal is the amplitude |x̂(ω)| of
1
1
the frequency ω part of the input signal multiplied by A(ω) = |−LCω2 +iRCω+1|
= √
.
2 2
2 2 2
(1−LCω ) +R C ω
A
ω
We shall shortly learn how to convert (6) into an equation that expresses the time domain output signal y(t)
in terms of the time domain input signal x(t).
d
Example 6 The Fourier transform, rect(ω),
of the rectangular pulse function of Example 2 decays rather
1
slowly, like ω for large ω. We can try suppressing large frequencies by eliminating the discontinuities at
t = ± 21 in rect(t). For example
g(t) =

0
if t ≤ − 85




5
5
3


 4(t + 8 ) if − 8 ≤ t ≤ − 8
1
if − 83 ≤ t ≤




4( 85 − t) if 38 ≤ t ≤ 85



0
if t ≥ 58
y
1
y = g(t)
3
8
− 58 − 38
3
8
5
8
t
It is not very difficult to evaluate the Fourier transform ĝ(ω) directly. But it easier to use properties (L)–(D),
since
y


if t ≤ − 85


0






5
3
if − 58 ≤ t ≤ − 83 


4

8
8
′
3
3
if − 8 ≤ t ≤ 8
g (t) = 0
= s4 (t) + s5 (t)
t


− 85 − 83


3
5




−4 if 8 ≤ t ≤ 8






5
y = g ′ (t)
0
if t ≥ 8
(
H = 4, W = 41 , C = − 12 for n = 4
where sn (t) = rHWC (t) with
for n = 5
H = −4, W = 41 , C = 12
By Example 3, the Fourier transform of
ŝ4 (ω) + ŝ5 (ω) =
c Joel Feldman.
2007. All rights reserved.
8 iω/2
ωe
sin ω8 − ω8 e−iω/2 sin ω8 = 2i ω8 sin ω2 sin ω8
March 1, 2007
The Fourier Transform
6
By Property (D), the Fourier transform of g ′ (t) is iω times ĝ(ω). So the Fourier transform of g(t) is
the Fourier transform of g ′ (t):
1
2i ω8 sin ω2 sin ω8 = ω162 sin ω2 sin ω8
ĝ(ω) = iω
1
iω
times
This looks somewhat like the the Fourier transform in Example 2 but exhibits faster decay for large ω.
1
ĝ(ω)
−2π
ω
2π
Parseval’s Relation
The energy carried by a signal f (t) is defined(2) to be
Z ∞
Z ∞
|f (t)|2 dt =
f (t)f (t) dt
−∞
−∞
We would like to express this energy in terms of the Fourier transform fˆ(ω). To do so, substitute in that
Z ∞
Z ∞
Z ∞
1
1
1
f (t) = 2π
fˆ(ω) eiωt dω = 2π
fˆ(ω)e−iωt dω
fˆ(ω)eiωt dω = 2π
−∞
This gives
Z
∞
−∞
2
|f (t)| dt =
1
2π
=
1
2π
This formula,
Duality
R∞
−∞
Z
−∞
∞
−∞
∞
Z
−∞
|f (t)|2 dt =
1
2π
Z
∞
−∞
f (t)fˆ(ω)e−iωt dω dt =
−∞
fˆ(ω) fˆ(ω) dω
R∞
−∞
=
1
2π
1
2π
Z
∞
−∞
∞
Z
−∞
fˆ(ω)
Z
∞
f (t)e
−iωt
−∞
dt dω
|fˆ(ω)|2 dω
|fˆ(ω)|2 dω, is called Parseval’s relation.
The duality property says that if we build a new time–domain function g(t) = fˆ(t) by exchanging the
roles of time and frequency, then the Fourier transform of g is
ĝ(ω) = 2πf (−ω)
(Du)
To verify this, just write down just the definition of ĝ(ω) and the Fourier inversion formula (4) for f (t) and,
in both integrals, make a change of variables so that the integration variable is s:
Z ∞
Z ∞
Z ∞
t=s
−iωt
−iωt
ˆ
ĝ(ω) =
g(t)e
dt =
f (t)e
dt
=
fˆ(s) e−iωs ds
−∞
−∞
−∞
Z ∞
Z ∞
ω=s 1
iωt
1
ˆ
f (ω)e
dω = 2π
fˆ(s) eist ds
f (t) = 2π
−∞
−∞
So ĝ(ω), which is given by the last integral of the first line is exactly 2π times the last integral of the second
line with t replaced by −ω, which is 2πf (−ω).
(2)
By way of motivation for this definition, imagine that f (t) is the voltage v(t) at time t across a resistor of resistance R.
1
Then the power (i.e. energy per unit time) dissipated by the resistor is v(t)i(t) = R
v(t)2 . So the total energy dissipated
R
1
2
by the resistor will be R v(t) dt, which, up to a constant, agrees with our definition.
c Joel Feldman.
2007. All rights reserved.
March 1, 2007
The Fourier Transform
7
Example 7 In this example, we shall compute the Fourier transform of sinc(t) = sint t . Our starting point
is Example 2, where we saw that the Fourier transform of the rectangular pulse rect(t) of height one and
d
d
width one is rect(ω)
= 2 sin ω = sinc ω . So, by duality (Du), with f (t) = rect(t) and g(t) = fˆ(t) = rect(t),
ω
2
2
the Fourier transform of g(t) = sinc 2t is ĝ(ω) = 2πf (−ω) = 2π rect(−ω) = 2π rect(ω), since rect(t) is
even. So we now know that the Fourier transform of sinc 2t is 2π rect(t). To find the Fourier transform
of sinc(t), we just need to scale the 12 out of 2t . So we apply the scaling property (S) with f (t) = sinc 2t
and h(t) = sinc(t) = f (2t) = f αt where α = 12 . By (S), the Fourier transform of h(t) = sinc(t) is
αfˆ(αω) = α 2π rect(αω) = π rect ω2 .
Multiplication and Convolution
A very common operation in signal processing is that of filtering. It is used to eliminate high frequency
noise and also to eliminate sixty cycle hum, arising from the ordinary household AC current. It is also
used to extract the signal from any one desired radio or TV station. In general, the filter is described
by a function Ĥ(ω) of frequency. For example you might have Ĥ(ω) = 1 for desired frequencies and
Ĥ(ω) = 0 for undesirable frequencies. When a signal f (t) is fed into the filter, an output signal g(t), whose
Fourier transform is fˆ(ω)Ĥ(ω) is produced. For example, the RLC circuit of Example 5 is a filter with
1
Ĥ(ω) = −LCω2 +iRCω+1
.
The question “what is g(t)?” remains. Of course it is the inverse Fourier transform
g(t) =
1
2π
Z
∞
fˆ(ω)Ĥ(ω)eiωt dω
−∞
of fˆ(ω)Ĥ(ω), but we would like to express it more directly in terms of the original time–domain signal f (t).
R∞
So let’s substitute fˆ(ω) = −∞ f (τ )e−iωτ dτ (which expresses fˆ(ω) in terms f (t)) and see if we can simplify
the result.
Z ∞Z ∞
g(t) =
=
=
1
2π
Z
∞
−∞
∞
Z
f (τ )e−iωτ Ĥ(ω)eiωt dτ dω
Z ∞
1
Ĥ(ω)eiω(t−τ ) dω dτ
f (τ ) 2π
−∞
−∞
−∞
−∞
f (τ )H(t − τ ) dτ
The last integral is called a convolution integral and is denoted
(f ∗ H)(t) =
Z
∞
−∞
f (τ )H(t − τ ) dτ
If we make the change of variables τ = t − τ ′ , dτ = −dτ ′ we see that we can also express
(f ∗ H)(t) =
Z
−∞
∞
′
′
′
f (t − τ )H(τ ) (−dτ ) =
Z
∞
−∞
f (t − τ ′ )H(τ ′ ) dτ ′
We conclude that
ĝ(ω) = fˆ(ω)Ĥ(ω) =⇒ g(t) = (f ∗ H)(t)
(C)
Example 8 The convolution integral gives a sort of moving weighted average, with the weighting determined
by H. Its value at time t involves the values of f for times near t. I say “sort of” because there is no
R∞
requirement that −∞ H(t) dt = 1 or even that H(t) ≥ 0.
c Joel Feldman.
2007. All rights reserved.
March 1, 2007
The Fourier Transform
8
Suppose for example, that we use the filter with
Ĥ(ω) =
2
ω
sin ω2
The graph of this function was given in Example 2. It is a “low pass filter” in the sense that it lets through
most small frequencies ω and suppresses high frequencies. It is not really a practical filter, but I have chosen
it anyway because it is relatively easy to see the effect of this filter in t–space. From Example 2 we know
that
1
1
H(t) = 1 if − 2 < t < 2
0 otherwise
So when this filter is applied to a signal f (t), the output at time t is
(f ∗ H)(t) =
=
Z
∞
−∞
Z
f (t − τ )H(τ ) dτ =
t+ 21
Z
1
2
− 12
f (t − τ ) dτ = −
Z
t− 21
f (τ ′ ) dτ ′
t+ 12
where τ ′ = t − τ, dτ ′ = −dτ
f (τ ′ ) dτ ′
t− 21
which is the area under the part of the graph of f (τ ′ ) from τ ′ = t − 12 to τ ′ = t + 21 . To be concrete, suppose
that the input signal is
y
n
′
f (τ ′ ) = 1 if τ ′ ≥ 0
0 if τ < 0
τ′
Then the output signal at time t is the area under the part of this graph from τ ′ = t − 21 to τ ′ = t + 21 .
◦ if t + 12 < 0, that is t < − 21 , then f (τ ′ ) = 0 for all t − 21 < τ ′ < t + 12 (see the figure below) so that
(f ∗ H)(t) = 0
y
t− 21
◦ if t +
1
2
1
2
τ′
t+ 12
≤ 0, that is − 21 ≤ t ≤ 21 , then f (τ ′ ) = 0 for t − 21 < τ ′ < 0 and f (τ ′ ) = 1 for
R t+ 1
R t+ 1
(see the figure below) so that (f ∗ H)(t) = t− 12 f (τ ′ ) dτ ′ = 0 2 1 dτ ′ = t + 21
≥ 0 but t −
0 < τ′ < t +
y = f (τ ′ )
1
2
2
y
t− 12
◦ if t −
R t+ 12
t− 21
1
2
y = f (τ ′ )
τ′
t+ 21
> 0, that is t > 21 , then f (τ ′ ) = 1 for all t −
1
2
< τ′ < t +
1
2
and (f ∗ H)(t) =
1 dτ ′ = 1.
y
R t+ 21
t− 21
f (τ ′ ) dτ ′ =
y = f (τ ′ )
t− 12
t+ 21
τ′
All together, the output signal is
c Joel Feldman.
2007. All rights reserved.
March 1, 2007
The Fourier Transform
9
y


0
(f ∗ H)(t) = t +


1
if t <
1
2
− 21
if − 21 ≤ t ≤
if t >
y = (f ∗ H)(t)
1
2
1
2
− 21
1
2
t
Example 9
For a similar, but more complicated example, we can through the procedure of Example 8 with H(t)
still being rect(t) but with f (t) replaced by the more complicated signal s(t) of Example 4. This time the
output signal at time t is
(s ∗ H)(t) =
Z
∞
−∞
s(t − τ )H(τ ) dτ =
Z
1
2
− 21
s(t − τ ) dτ =
Z
t+ 12
s(τ ′ ) dτ ′
t− 21
where τ ′ = t − τ
which is the area under the part of the graph of s(τ ′ ) from τ ′ = t − 12 to τ ′ = t + 12 . We can read off this
area from the figures below. In each figure, the left hand vertical dotted line is τ ′ = t − 21 and the right hand
vertical dotted line is τ ′ = t + 12 . The first figure has t so small that the entire interval t − 12 < τ ′ < t + 21 is to
the left of τ ′ = −2, where the graph of s(τ ′ ) “starts”. Then, in each successive figure, we increase t, moving
the interval to the right. In the final figure t is large enough that the entire interval t − 21 < τ ′ < t + 21 is to
the right of τ ′ = 4, where the graph of s(τ ′ ) “ends”.
y
2
y
2
y = s(τ ′ )
1
t− 21
t+ 12
1
τ′
−2 −1
0
1
2
3
4
Case: t + 21 < −2, that is t < − 52
area = 0
y
2
τ′
−2 −1
0
1
2
3
4
Case: −2 ≤ t + 21 ≤ −1, that is − 25 ≤ t ≤ − 23
area = 2((t + 21 ) − (−2)) = 2t + 5
y
2
y = s(τ ′ )
1
1
−2 −1
0
1
2
3
4
3
1
Case: −1 ≤ t + 2 ≤ 0, that is − 2 ≤ t ≤ − 21
area = 2((−1) − (t − 12 )) = −1 − 2t
τ′
−2 −1
0
1
2
Case: 1 ≤ t + 21 ≤ 2, that is
area = 1
2007. All rights reserved.
3
4
1
3
≤
t
≤
2
2
τ′
1
2
y
2
y = s(τ ′ )
1
y = s(τ ′ )
−2 −1
0
1
2
3
4
1
1
Case: 0 ≤ t + 2 ≤ 1, that is − 2 ≤ t ≤
area = t + 21
y
2
c Joel Feldman.
y = s(τ ′ )
1
τ′
τ′
−2 −1
0
1
2
3
4
Case: 2 ≤ t + 21 ≤ 3, that is 23 ≤ t ≤ 25
area = [2 − (t − 12 ) + 12 [(t + 21 ) − 2] = 47 − 21 t
March 1, 2007
The Fourier Transform
10
y
2
y
2
1
1
−2 −1
0
1
2
1
Case: 3 ≤ t + 2 ≤ 4, that is
area = 12
τ′
3
4
5
7
2 ≤ t≤ 2
−2 −1
0
1
2
3
4
1
7
Case: 4 ≤ t + 2 ≤ 5, that is 2 ≤ t ≤ 29
area = 21 [4 − (t − 12 )] = 49 − 21 t
τ′
y
2
y = s(τ ′ )
1
−2
−1
0
1
2
3
Case: t − 12 > 4, that is t >
t− 21
4
t+ 12
9
2
area = 0
All together, the graph of the output signal is
(s ∗ H)(t)
2
1
−2
−1
0
1
2
3
4
t
Impulses
The inverse Fourier transform H(t) of Ĥ(ω) is called the impulse response function of the filter, because
it is the output generated when the input is an impulse at time 0. An impulse, usually denoted δ(t) (and
called a “delta function”) takes the value 0 for all times t 6= 0 and the value ∞ at time t = 0. In fact it is
so infinite at time 0 that the area under its graph is exactly 1. Of course there isn’t any such function, in
∞
y
y = δ(t)
t
the usual sense of the word. But it is possible to generalize the concept of a function (to something called
a distribution or a generalized function) so as to accommodate delta functions. One generalization involves
c Joel Feldman.
2007. All rights reserved.
March 1, 2007
The Fourier Transform
11
taking the limit as ε → 0 of “approximate delta functions” like
δε (t) =
1
2ε
0
y
1
2ε
if −ε < t < ε
y = δε (t)
otherwise
−ε ε
t
A treatment of these generalization procedures is well beyond the scope of this course. Fortunately, in practice
R∞
it suffices to be able to compute the value of the integral −∞ δ(t)f (t) dt for any continuous function f (t)
and that is easy. Because δ(t) = 0 for all t 6= 0, δ(t)f (t) is the same as δ(t)f (0). (Both are zero for t 6= 0.)
Because f (0) is a constant, the area under δ(t)f (0) is f (0) times the area under δ(t), which we already said
is 1. So
Z ∞
δ(t)f (t) dt = f (0)
−∞
In particular, choosing f (t) = e−iωt gives the Fourier transform of δ(t):
Z ∞
δ̂(ω) =
δ(t)e−iωt dt = e−iωt t=0 = 1
−∞
By the time shifting property (T), the Fourier transform of δ(t − t0 ) (where t0 is a constant) is e−iωt0 . We
may come to the same conclusion by first making the change of variables τ = t − t0 to give
Z ∞
Z ∞
τ =t−t0
−iωt
δ(τ )e−iω(τ +t0 ) dτ
δ(t − t0 )e
dt =
−∞
−∞
R∞
and then applying −∞ δ(τ )f (τ ) dτ = f (0) with f (τ ) = e−iω(τ +t0 ) .
Returning to the impulse response function, we can now verify that the output generated by a filter
Ĥ(ω) in response to the impulse input signal δ(t) is indeed
Z ∞
(δ ∗ H)(t) =
δ(τ )H(t − τ ) dτ = H(t)
where we have applied
R∞
−∞
−∞
δ(τ )f (τ ) dτ = f (0) with f (τ ) = H(t − τ ).
Example 10 We saw in (6) that for an RLC circuit
Ĥ(ω) =
1
−LCω 2 +iRCω+1
To make the numbers work out cleanly, let’s choose R = 1, L = 6 and C = 5. Then
Ĥ(ω) =
1
−6ω 2 +i5ω+1
=
1
(3iω+1)(2iω+1)
Recall from Example 1 that
f (t) =
e−at
0
if t ≥ 0
if t < 0
⇒
fˆ(ω) =
1
a+iω
(7)
So we can determine the impulse response function H(t) for the RLC filter just by using partial fractions to
1
’s. Since
write H(ω) as a linear combination of a+iω
a
b
a(2iω + 1) + b(3iω + 1)
(2a + 3b)iω + (a + b)
1
=
+
=
=
(3iω + 1)(2iω + 1)
3iω + 1 2iω + 1
(3iω + 1)(2iω + 1)
(3iω + 1)(2iω + 1)
⇐⇒ 2a + 3b = 0, a + b = 1 ⇐⇒ b = 1 − a, 2a + 3(1 − a) = 0 ⇐⇒ a = 3, b = −2
Ĥ(ω) =
c Joel Feldman.
2007. All rights reserved.
March 1, 2007
The Fourier Transform
12
we have, using (7) with a =
1
3
and a = 21 ,
3
2
Ĥ(ω) =
−
=
3iω + 1 2iω + 1
1
3
1
−
+ iω
1
2
1
+ iω
⇒
H(t) =
e−t/3 − e−t/2
0
if t ≥ 0
if t < 0
which has graph
y
y = H(t)
t
Example 10 (again) Here is a sneakier way to do the partial fraction expansion of Example 10. We know
that Ĥ(ω) has a partial fraction expansion of the form
Ĥ(ω) =
1
a
b
=
+
(3iω + 1)(2iω + 1)
3iω + 1 2iω + 1
The fast way to determine a is to multiply both sides of this equation by (3iω + 1)
b(3iω + 1)
1
=a+
2iω + 1
2iω + 1
and then evaluate both sides at 3iω + 1 = 0. Then the b term becomes zero, because of the factor (3iω + 1)
in the numerator and we get
1 1
a=
= 1 =3
2iω + 1 iω=− 1
3
3
Similarly, multiplying by (2iω + 1) rather than (3iω + 1),
1 1
1 =
= 1 = −2
b=
3iω + 1 2iω+1=0
3iω + 1 iω=− 1
−2
2
c Joel Feldman.
2007. All rights reserved.
March 1, 2007
The Fourier Transform
13
Properties of the Fourier Transform
Property
Signal
x(t) =
1
2π
y(t) =
1
2π
R∞
−∞
R∞
−∞
Linearity
Ax(t) + By(t)
Time shifting
x(t − t0 )
Frequency shifting
Scaling
Time shift & scaling
Fourier Transform
R∞
x̂(ω) = −∞ x(t)e−iωt dt
R∞
ŷ(ω) = −∞ y(t)e−iωt dt
x̂(ω)eiωt dω
ŷ(ω)eiωt dω
Ax̂(ω) + B ŷ(ω)
e−iωt0 x̂(ω)
eiω0 t x(t)
x αt
x
x̂(ω − ω0 )
|α|x̂(αω)
t−t0
α )
iω0 t
|α|e−iωt0 x̂(αω)
0
x̂ ω−ω
α
Frequency shift & scaling
|α|e
Conjugation
x(t)
x̂(−ω)
Time reversal
x(−t)
x̂(−ω)
t–Differentiation
x(αt)
′
x (t)
iω x̂(ω)
x(n) (t)
(iω)n x̂(ω)
ω–Differentiation
tx(t)
Convolution
Multiplication
tn x(t)
R∞
−∞ x(τ )y(t − τ ) dτ
d
x̂(ω)
i dω
n
d
i dω x̂(ω)
Duality
x̂(t)
x̂(ω)ŷ(ω)
R∞
1
2π −∞ x̂(θ)ŷ(ω − θ) dθ
x(t)y(t)
2πx(−ω)
R∞
R∞
1
|x(t)|2 dt = 2π
|x̂(ω)|2 dω
−∞
−∞
Parseval
n
0
if t < 0,
e−at if t > 0
−at
if t < 0,
−at
e u(−t) = e
0
if t > 0
e−at u(t) =
e−a|t|
(
Boxcar in time
rect(t) =
General boxcar
rHWC (t) =
Boxcar in frequency
1
2π
Impulse in time
δ(t − t0 )
Single frequency
c Joel Feldman.
sinc
if |t| <
1
2
1
2
0 if |t| >
(
H if |t − C| <
t
2
1
0
=
if |t − C| >
1
πt
sin
t
2
1
(a constant, Re a < 0)
a + iω
2a
(a constant, Re a > 0)
2
a + ω2
−
sinc
W
2
W
2
ω
2
=
2
ω
sin
HW e−iωC sinc
rect(ω) =
1
0
ω
2
Wω
2
Wω
= e−iωC 2H
ω sin 2
if |ω| < 1/2
if |ω| > 1/2
e−iωt0
e−iωt0 x(t0 )
δ(t − t0 ) x(t)
eiω0 t
2007. All rights reserved.
1
(a constant, Re a > 0)
a + iω
2π δ(ω − ω0 )
March 1, 2007
The Fourier Transform
14